Chapter 10.3, Problem 40E

To determine

Solve the system of linear equations.

Answer to Problem 40E

The solution for given system is $\left(-1,3,5\right)$ .

Explanation of Solution

Given:

Equation given,

  $\begin{array}{l}2x-3y+5z=14\\ 4x-y-2z=-17\\ -x-y+z=13\end{array}$

Concept Used:

The concept of Gaussian elimination to solve the system of linear equation’s.

Calculation:

Consider first the given equations,

  $\begin{array}{l}2x-3y+5z=14\\ 4x-y-2z=-17\\ -x-y+z=13\end{array}$

Step $1$ :

In the first step first convert the given, equation into augmented matrix.

  $\left[\begin{array}{cccc}2& -3& 5& 14\\ 4& -1& -2& -17\\ -1& -1& 1& 3\end{array}\right]$

Step $2$ :

In the second step convert the augmented matrix and use the Gauss Elimination method.

For this row operation is performed. In the first row operation,

  $R_1\leftrightarrow R_3$

  $R_1\leftrightarrow R_3=$ $=\left[\begin{array}{cccc}-1& -1& 1& 3\\ 4& -1& -2& -17\\ 2& -3& 5& 14\end{array}\right]$

The next row operation is,

  $\left(-1\right)+4R_1+2R_1$

  $=\left[\begin{array}{cccc}1& 1& -1& -3\\ 0& -5& 2& -5\\ 0& -5& 7& 20\end{array}\right]$

Now perform,

  $-R_2$

  $=\left[\begin{array}{cccc}1& 1& -1& -3\\ 0& -5& 2& -5\\ 0& 0& 1& 5\end{array}\right]$

Step3:

In the Step3 perform the following operations on the matrix,

Divide $\frac{R_3}{5}$ ,

  $\frac{R_3}{5}$

  $=\left[\begin{array}{cccc}1& 1& -1& -3\\ 0& -5& 2& -5\\ 0& 0& 1& 5\end{array}\right]$

The next row operation is $R_3-2R_1$ ,

  $R_3-2R_1$

  $=\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& -5& 0& -15\\ 0& 0& 1& 5\end{array}\right]$

Now the last matrix operation to be performed is,

  $\frac{1}{5}{R}_2÷\left(-5\right)=\left[\begin{array}{cccc}1& 0& 0& -1\\ 0& 1& 0& 3\\ 0& 0& 1& 5\end{array}\right]$

Thus the above equation is obtained .

  $\begin{array}{l}x=-1\\ y=3\\ z=5\end{array}$

Conclusion:

Hence, the solution for given system is $\left(-1,3,5\right)$ .